Find the area of

Question:

Find the area of $\triangle A B C$ with vertices $A(0,-1), B(2,1)$ and $C(0,3)$. Also, find the area of the triangle formed

by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1.  

 

Solution:

Let A(x1 = 0, y1 = −1), B(x2 = 2, y2 = 1) and C(x3 = 0, y3 = 3) be the given points. Then

Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$=\frac{1}{2}[0(1-3)+2(3+1)+0(-1-1)]$

$=\frac{1}{2} \times 8=4$ sq. units

So, the area of the triangle $\Delta A B C$ is 4 sq. units.

Let D(a1b1), E(a2b2) and F(a3b3) be the midpoints of ABBC and AC respectively. Then

$a_{1}=\frac{0+2}{2}=1 \quad b_{1}=\frac{-1+1}{2}=0$

$a_{2}=\frac{2+0}{2}=1 \quad b_{2}=\frac{1+3}{2}=2$

$a_{3}=\frac{0+0}{2}=0 \quad b_{3}=\frac{-1+3}{2}=1$

Thus, the coordinates of D, E and F are D(a1 = 1, b1 = 0), E(a2 = 1, b2 = 2) and F(a3 = 0, b3 = 1). Now

Area $(\Delta D E F)=\frac{1}{2}\left[a_{1}\left(b_{2}-b_{3}\right)+a_{2}\left(b_{3}-b_{1}\right)+a_{3}\left(b_{1}-b_{2}\right)\right]$

$=\frac{1}{2}[1(2-1)+1(1-0)+0(0-2)]$

$=\frac{1}{2}[1+1+0]=1$ sq. unit

So, the area of the triangle $\Delta D E F$ is 1 sq. unit.

Hence, $\Delta A B C: \Delta D E F=4: 1$

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